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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 22:24:01 IST
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A shell is fired from a gun from the bottom of the hill along it's slope.The slope of the hill is 30 degrees,and angle of the barrel is 60 degrees.The initial velocity u of the shell is 21m/s.Find the distance along the slope from the gun to the point at which shell falls,take g=10m/ssquare
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You can never solve a problem on the level on which it was created.
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29 Nov 2007 22:50:13 IST
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is angle 60 degree is with slope or with horizontal
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LOKESH SARDANA,
department of Production and Industrial engineering,
Indian Institute of Technology,Roorkee.
Happiness can be found, even in the darkest of times, if one only remembers to turn on the light.
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let angle of slope = @
let angle of projection = #
then range is
=
-2(u)^2cos#sin(# - @)
---------------------------
g cos@
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29 Nov 2007 22:57:54 IST
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eqn of projectile motion y=xtanw-gx^2/2u^2cos^2w (w is angle o pr)
putting g=10, u=21, w=60 dg
y=rt(3)x-20x^2/441 (i)
eqn of hill's slope is y=xtan30 or y=x/rt(3) (ii)
solving eqns (i) n (ii)
x=441/10rt(3) and y=441/30 coordinates of point where shell falls
dist frm hills bottom = rt(x^2+y^2)
=441/15
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29 Nov 2007 22:59:38 IST
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Is it 14.7 m?
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my future signature:
LOKESH SARDANA,
department of Production and Industrial engineering,
Indian Institute of Technology,Roorkee.
Happiness can be found, even in the darkest of times, if one only remembers to turn on the light.
Albus Dumbledore
Harry Potter and the Prisoner of Azkaban movie
There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle.
-- Albert Einstein
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29 Nov 2007 23:04:42 IST
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rhd is right
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Anyone who has never made a mistake has never tried anything new
You can never solve a problem on the level on which it was created.
When you are courting a nice girl an hour seems like a second. When you sit on a red-hot cinder a second seems like an hour. That's relativity
Purpose of solving a problem is not simply to get the answer(the answer is only an evidence) but to develop your thinking ability |
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29 Nov 2007 23:09:54 IST
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Simple question really... consider the slope of the hill to be as x axis ( tilt the axes)
now in normal equations of projectile motion. substitute g for gcos@ where @ is the angle of the barrel wrt the slope. and of course where theta comes normally u have to take the angle between gun and slope...
in this question u need range so apply range formula.
R = u^2. sin2@ / g
angle between gun and slope is 30 degrees = @
g--> g cos @
So
R = (21)^2 . sin60 / g cos@
R = 441 root(3) / (2 * 10 * root(3) / 2 )
R = 441 / 10
R = 44.1 m along the slope
see how easy it is now :D
cheers!
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30 Nov 2007 07:58:04 IST
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Sorry Spidey u r wrong,I also got the same answer,but rhd's answer is rite
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Anyone who has never made a mistake has never tried anything new
You can never solve a problem on the level on which it was created.
When you are courting a nice girl an hour seems like a second. When you sit on a red-hot cinder a second seems like an hour. That's relativity
Purpose of solving a problem is not simply to get the answer(the answer is only an evidence) but to develop your thinking ability |
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30 Nov 2007 12:58:45 IST
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if u've not got it, here's it is: consider the foot of hill on gun's side as origin slope of hill's surface = 30deg so its eqn is y=xtan30 or y=x/ [ ] 3 ....(i) now consider the trajectory of projectile its eqn is y=xtan  - gx 2/2u 2cos 2 now put the values of g=10, u=21, =60 deg (the angle of projection is 60 deg) the eqn becomes y= 3x-20x 2/441 ....(ii) the point on the hill where the shell will fall will be the intersection of eqns (i) and (ii) solving both, we get x=441/10 3 and y=441/30 m so its distance frm the origin = (x 2+y 2) which u get as 441/15 m
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
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<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
I am only one,
But still I am one.
I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
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